Consider $x^{2}-2x-80$. Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-80$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-2$$ $$ab=1\left(-80\right)=-80$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-80$.
Rewrite $x^{2}-2x-80$ as $\left(x^{2}-10x\right)+\left(8x-80\right)$.
$$\left(x^{2}-10x\right)+\left(8x-80\right)$$
Factor out $x$ in the first and $8$ in the second group.
$$x\left(x-10\right)+8\left(x-10\right)$$
Factor out common term $x-10$ by using distributive property.
$$\left(x-10\right)\left(x+8\right)$$
Rewrite the complete factored expression.
$$2\left(x-10\right)\left(x+8\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$2x^{2}-4x-160=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
